Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{n + 9}{n^2 - 10n} \div \dfrac{2n - 16}{n^2 - 18n + 80} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{n + 9}{n^2 - 10n} \times \dfrac{n^2 - 18n + 80}{2n - 16} $ First factor the quadratic. $r = \dfrac{n + 9}{n^2 - 10n} \times \dfrac{(n - 10)(n - 8)}{2n - 16} $ Then factor out any other terms. $r = \dfrac{n + 9}{n(n - 10)} \times \dfrac{(n - 10)(n - 8)}{2(n - 8)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (n + 9) \times (n - 10)(n - 8) } { n(n - 10) \times 2(n - 8) } $ $r = \dfrac{ (n + 9)(n - 10)(n - 8)}{ 2n(n - 10)(n - 8)} $ Notice that $(n - 8)$ and $(n - 10)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ (n + 9)\cancel{(n - 10)}(n - 8)}{ 2n\cancel{(n - 10)}(n - 8)} $ We are dividing by $n - 10$ , so $n - 10 \neq 0$ Therefore, $n \neq 10$ $r = \dfrac{ (n + 9)\cancel{(n - 10)}\cancel{(n - 8)}}{ 2n\cancel{(n - 10)}\cancel{(n - 8)}} $ We are dividing by $n - 8$ , so $n - 8 \neq 0$ Therefore, $n \neq 8$ $r = \dfrac{n + 9}{2n} ; \space n \neq 10 ; \space n \neq 8 $